Entanglement Bounds in the XXZ Quantum Spin Chain

Abstract

We consider the XXZ spin chain, characterized by an anisotropy parameter $\Delta >1$, and normalized such that the ground state energy is 0 and the ground state given by the all-spins-up state. The energies $E_K = K(1-1/\Delta )$, $K=1,2,\ldots$, can be interpreted as K-cluster breakup thresholds for down-spin configurations. We show that, for every K, the bipartite entanglement of all states with energy below the $(K+1)$-cluster breakup satisfies a log-corrected area law. This generalizes a result by Beaud and Warzel, who considered energies in the droplet spectrum (i.e., below the 2-cluster breakup). For general K, we find an upper logarithmic bound with pre-factor $2K-1$. We show that this constant is optimal in the Ising limit $\Delta =\infty$. Beaud and Warzel also showed that after introducing a random field and disorder averaging the log-corrected area law becomes a strict area law, again for states in the droplet regime. For the Ising limit with random field, we show that this result does not extend beyond the droplet regime. Instead, we find states with energies of an arbitrarily small amount above the K-cluster breakup whose entanglement satisfies a logarithmically growing lower bound with pre-factor $K-1$.

Auxiliary Results

Here, we collect some technical lemmas and their proofs. We strongly recommend to draw many pictures while reading these proofs.

1.1 On the Distance Formula for the Case of the Chain

If the underlying graph is the chain or a subinterval thereof, when dealing with its n-th symmetric power, it is natural to only consider orderedn-element subsets of the form $X=\{x_1<x_2<\cdots<x_n\}, Y=\{y_1<y_2<\cdots <y_n\}$. In particular, the ordered labeling used here indeed gives the minimizer in

$$\begin{aligned} d_n(X,Y)=\min _{\pi \in \mathfrak {S}_n} \sum _{j=1}^n |x_j - y_{\pi (j)}|=\sum _{j=1}^n |x_j - y_{j}|\,, \end{aligned}$$

(A.1)

the general distance formula on symmetric product graphs, see (2.8). Indeed, one sees this by the following argument:

• (i) Consider first the case $n=2$. Thus, let $x_1<x_2$ and $y_1 < y_2$. It follows by inspection (of the possible relative locations of $x_1$, $x_2$ to $y_1$, $y_2$) that for all possible cases

  $$\begin{aligned} |x_1 - y_1| + |x_2-y_2| \le |x_1-y_2| + |x_2-y_1|. \end{aligned}$$

  (A.2)

• (ii) In the general case, let $x_1< x_2< \cdots < x_n$ and $y_1< y_2< \cdots < y_n$ be two ordered configurations. Let $\pi \in \mathfrak {S}_n$ not be the identity. Thus, there is at least one k such that $y_{\pi (k+1)} < y_{\pi (k)}$. Therefore, by (i),

  $$\begin{aligned} \sum _j |x_j - y_{\pi (j)}|= & {} \sum _{j\not \in \{k,k+1\}} |x_j- y_{\pi (j)}| + |x_k-y_{\pi (k)}| + |x_{k+1}- y_{\pi (k+1)}| \nonumber \\\ge & {} \sum _{j\not \in \{k,k+1\}} |x_j- y_{\pi (j)}| + |x_k - y_{\pi (k+1)}| + |x_{k+1}- y_{\pi (k)}|.\nonumber \\ \end{aligned}$$

  (A.3)

  After finitely many transpositions of this type, one arrives back at the ordered configuration $(y_1,\ldots ,y_n)$, proving (A.1).

1.2 Closest Droplets and K-Cluster Configurations

We will generally label n-particle configurations $W\in \mathcal {V}_n$ in increasing order $W=\{w_1< w_2< \cdots < w_n\}$.

Lemma A.1

Let $W \in \mathcal {V}_n$.

• (i) If n is odd, then the unique n-droplet $\hat{W}=\{\hat{w}_1<\cdots <\hat{w}_n\}$ closest to W with respect to $d_n(\cdot ,\cdot )$ is the droplet which shares its central particle with W, i.e., $w_{(n+1)/2} = \hat{w}_{(n+1)/2}$.

• (ii) If n is even, then there are $w_{\frac{n}{2}+1} - w_{\frac{n}{2}}+1$ (the size of the “central gap” in W plus one) closest n-droplets, with central particles at $w_{\frac{n}{2}} , \ldots , w_{\frac{n}{2}+1}$ (the positions in the central gap and its boundary), respectively.

Proof

This follows from elementary considerations, which can be based on the following facts: If $W \in \mathcal {V}_n$, $V\in \mathcal {V}_{n,1}$ and $V\pm 1 = \{v_1 \pm 1, \ldots , v_n \pm 1\}$ denote the right and left shift of V (if they are still in $\mathcal {V}_n$), then

$$\begin{aligned} |\{j: w_j< v_j\}|< & {} |\{j:w_j \ge v_j\}| \quad \Longrightarrow \quad d_n(V-1,W) > d_n(V,W). \qquad \qquad \qquad \qquad \end{aligned}$$

(A.4)

$$\begin{aligned} |\{j: w_j < v_j\}|= & {} |\{j:w_j \ge v_j\}| \quad \Longrightarrow \quad d_n(V-1,W) = d_n(V,W). \qquad \qquad \qquad \qquad \end{aligned}$$

(A.5)

$$\begin{aligned} |\{j: w_j \le v_j\}|> & {} |\{j:w_j> v_j\}| \quad \Longrightarrow \quad d_n(V+1,W) > d_n(V,W). \qquad \qquad \qquad \qquad \end{aligned}$$

(A.6)

$$\begin{aligned} |\{j: w_j \le v_j\}|= & {} |\{j:w_j > v_j\}| \quad \Longrightarrow \quad d_n(V+1,W) = d_n(V,W). \end{aligned}$$

(A.7)

All of these properties follow by comparing how many $v_j$ get closer to $w_j$ or more distant from $w_j$ (in each case by one) under the corresponding shift. The last two properties are “mirroring” the first two.

To see (i), one can use the first and third property to see that starting from the droplet which shares the same center particle with W will always increase the $d_n$-distance. For the even case (ii), one sees the additional degeneracy via the second and fourth property, before the distance starts increasing under further shifting. $\square$

We say that an n-particle droplet $\hat{X}$ is centered at $\hat{x}$, or $\hat{x}$ is a central particle of the droplet $\hat{X}$, if it is of the form:

$$\begin{aligned} \hat{X}=\left\{ \hat{x}-\lfloor \frac{n}{2}\rfloor , \ldots ,\hat{x}-1,\hat{x},\hat{x}+1,\ldots , \hat{x}+\lfloor \frac{n}{2}\rfloor -\delta _{n,\text {even}}\right\} \end{aligned}$$

(A.8)

where

$$\begin{aligned} \delta _{n,\text {even}}=\left\{ \begin{array}{ll} 1 &{}\quad \hbox {if }n\hbox { is even} \\ 0 &{}\quad \hbox {if }n\hbox { is odd.} \end{array} \right. \end{aligned}$$

(A.9)

This means that $\hat{x}$ is the middle particle of $\hat{X}$ if n is odd, and for technical reasons we choose the define the central particle to be the bigger between the two middle particles when n is even.

Recall the definitions of the sets $\mathcal {V}_{n,k}$, $\mathcal {V}^=_{n,k}$ and $\Xi _{n,k}$ in (6.4), (6.6) and (6.7).

Lemma A.2

Let $X = \{x_1< \cdots < x_n\} \in \Xi _{n,k}$. Then, there exists $\hat{X} = \{ \hat{x}_1< \cdots < \hat{x}_n\} \in \mathcal {V}_{n,k}^=$ with $d_n(X,\mathcal {V}_{n,k}^=) = d_n(X,\hat{X})$ and $x_{j_r} = \hat{x}_{j_r}$, $r=1,\ldots , k$, where $\hat{x}_{j_r}$ is the central particle of the r-th cluster in $\hat{X}$ (written in increasing order).

We note that one may think of the $x_{j_r}$ as “magnets” within X. The closest k-cluster $\hat{X}$ is found by moving the other particles in X into clusters centered at the $x_{j_r}$.

Proof

By definition, there is an $\tilde{X}\in \mathcal {V}_{n,k}^=$ such that

$$\begin{aligned} d_n(X,\tilde{X})=\min _{Y\in \mathcal {V}_{n,k}^=}d_n(X,Y). \end{aligned}$$

(A.10)

We will show that $\tilde{X}$ is of the form required for $\hat{X}$ or can be slightly modified to yield the required form.

There exist $m_1,m_2,\ldots ,m_{k}\in \mathbb {N}$ with $\sum m_j=n$ such that $\tilde{X}$ can be written as a union of (non-touching) $m_r$-particle droplets $\tilde{X}_{r}$, i.e.,

$$\begin{aligned} \tilde{X}=\bigcup _{r=1}^k \tilde{X}_r, \end{aligned}$$

(A.11)

and we may choose the droplets $\tilde{X}_r$ as ordered in the sense that

$$\begin{aligned} \max \tilde{X}_r<\min \tilde{X}_{r+1}-1, \text { for all } 1\le r\le k-1. \end{aligned}$$

(A.12)

We consider a corresponding partition of X into k ordered subsets $\{X_r\}_{r=1}^k$ each containing $m_r$ particles from X. In particular, for each $r\in \{1,\ldots ,k\}$, we define $X_r\subseteq X$ as follows

$$\begin{aligned} X_1:= & {} \{x_1,\ldots ,x_{m_1}\},\ X_2:=\{x_{m_1+1},\ldots ,x_{m_1+m_2}\}, \ldots , X_k\nonumber \\:= & {} \{x_{n-m_r-1},\ldots ,x_n\}. \end{aligned}$$

(A.13)

Then, by Appendix A.1,

$$\begin{aligned} d_n(X,\tilde{X})= \sum _{r=1}^k d_{m_r}(X_r,\tilde{X}_r). \end{aligned}$$

(A.14)

We have $\min X_r \le \min \tilde{X}_r$ and $\max X_r \ge \max \tilde{X}_r$ for all r (otherwise the distance of $\tilde{X}$ to X could be reduced by shifting some of its clusters, contradicting the minimality property of $\tilde{X}$).

For each $r\in \{1,\ldots ,k\}$, we consider two cases and in each case argue with Lemma A.1 and the minimality property of $\tilde{X}$:

• (i) $m_r$ is odd. In this case, $X_j$ and $\tilde{X}_j$ must have their central particle in the same position.

• (ii) $m_r$ is even. Then, either $\tilde{X}_r$ or a suitable right-shift of $\tilde{X}_r$ will have the same central particle as $X_r$ (without change of the distance to $X_r$).

Accordingly modifying $\tilde{X}$, we get $\hat{X}$ of the required form.

We label the droplets of $\hat{X}$ and their corresponding configurations in $X\in \mathcal {V}_{n,k}^=$ using these central particles, i.e., we write

$$\begin{aligned} X=\bigcup _{r=1}^k X_{\hat{x}_r}, \text { and } \hat{X}=\bigcup _{r=1}^k \mathcal {C}_{\hat{x}_r},\ \mathcal {C}_{\hat{x}_r}\in \mathcal {V}_{m_r,1}^=, \text { and }\sum _{r=1}^k m_r=n. \end{aligned}$$

(A.15)

1.3 The Distances $d_{|Y|+k}(\mathcal {A}_{Y,k}, \mathcal {V}_{|Y|+k,K})$

Finally, we prove the properties of the distances $d_{|Y|+k}(\mathcal {A}_{Y,k}, \mathcal {V}_{|Y|+k,K})$ which we have used in the proof of Proposition 5.2 above.

Lemma A.3

To any $Y\subseteq \Lambda _0$ with $1\le |Y| \le \ell -1$ and $1\le k \le L-|Y|$, let

$$\begin{aligned} Y^{(k)} := Y \cup \{\ell +1, \ell +2, \ldots , \ell +k\} \in \mathcal {A}_{Y,k}. \end{aligned}$$

(A.16)

Then, for any $K\in \mathbb {N}$,

$$\begin{aligned} d_{|Y|+k}(\mathcal {A}_{Y,k}, \mathcal {V}_{|Y|+k,K}) = d_{|Y|+k}(Y^{(k)}, \mathcal {V}_{|Y|+k,K}). \end{aligned}$$

(A.17)

Proof of Lemma A.3

Let $V \in \mathcal {A}_{Y,k}$ and $W\in \mathcal {V}_{|Y|+k,K}$ a distance minimizing pair, i.e., $d_{|Y|+k}(V,W) = d_{|Y|+k}(\mathcal {A}_{Y,k}, \mathcal {V}_{|Y|+k,K})$. If $V\cap \{\ell +1,\ldots ,\ell +k\} \not = \{\ell +1,\ldots ,\ell +k\}$, then one can iteratively use Lemma A.4 below to move the right-most components of V to the left, until one arrives at $Y^{(k)}$, without increasing the distance to $\mathcal {V}_{|Y|+k,K}$. This gives (A.17). $\square$

Lemma A.4

Let $V = (v_1,\ldots , v_n) \in \mathcal {V}_n {\setminus } \mathcal {V}_{n,1}$ and $(v_{j+1},\ldots ,v_n)$ its right-most connected component. Let $V' = (v_1,\ldots ,v_j, v_{j+1}-1,\ldots , v_n-1)$. Then, for any $K\in \mathbb {N}$,

$$\begin{aligned} d_n(V', \mathcal {V}_{n,K}) \le d_n(V,\mathcal {V}_{n,K}). \end{aligned}$$

(A.18)

Proof

Choose $Z = (z_1,\ldots ,z_n) \in \mathcal {V}_{n,K}$ with minimal distance to V. It is easy to see that $v_1 \le z_1$ and $z_n \le v_n$. Write $Z= Z_1 \cup Z_2$, where $Z_2= (z_{k+1},\ldots ,z_n)$ is the right-most connected component of Z. Let $Z' := Z_1 \cup (Z_2-1) \in \mathcal {V}_{n,K}$. One can see that $k\le j$ ($k>j$ would be a contradiction to Z having minimal distance to V, as in this case one could move $z_{j+1}, \ldots , z_k$ to the $k-j$ sites directly to the left of $z_{k+1}$, reducing the distance to V without increasing the number of clusters of Z).

Now there are two cases: If $z_n<v_n$, then $d_n(V',Z) < d_n(V,Z)$. In case $z_n=v_n$, using $k\le j$, it follows that $d_n(V',Z') \le d_n(V,Z)$. Thus, in both cases, one gets (A.18). $\square$

Lemma A.5

The minimum of the numbers $d_{|Y|+k}(\mathcal {A}_{Y,k}, \mathcal {V}_{|Y|+k,K})$, $1\le k \le L-|Y|$, is attained for $k=1$.

Proof

By Lemma A.3, we have to consider $d_{|Y|+k}(Y^{(k)}, \mathcal {V}_{|Y|+k,K})$, $1\le k \le L-|Y|$. It is geometrically quite evident that these numbers are non-decreasing in k: If one can move $Y^{(k+1)}$ to a set Z with no more than K-clusters in s steps, then in doing so one has also moved $Y^{(k)}$ (appearing as a “restriction” of $Y^{(k+1)}$) to a K-cluster set (Z without its right-most element) in at most s steps. $\square$